Question

A cannon tilted up at a 29.0° angle fires a cannon ball at 81.0 m/s from atop a 22.0 m -high fortress wall. What is the ball's impact speed on the ground below?

Answer 1

Answer:

The speed of the ball when it hits the ground is 83.4 m/s

Explanation:

Please see the attached figure for a description of the problem.

The vector velocity at time "t" can be written as follows:

$v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)$

where:

v0 : module of the initial velocity vector.

α: launching angle.

g: acceleration due to gravity.

t: time

To calculate the impact speed, we can use this equation once we know the time at which the ball hits the ground. For this, we can use the equation for position:

$r = ( x0 + v0*t*cos\alpha ; y0 + v0*t*sin\alpha + 1/2 g*t^{2})$

where:

r= vector position

x0 = horizontal initial position

y0 = vertical initial position

The problem gives us information about the vertical displacement. If we take the base of the wall as the center of the reference system, at the time at which the ball hits the ground, the module of the vertical component of the vector position, r, is 0 m (see figure). The initial vertical position is  22 m.

if ry is the vertical component of the vector r at final time:

$ry = (0; y0 + v0*t*sin\alpha +1/2 g*t^{2})$

then, the module of ry is (see figure):

module ry =$0 = 22m + v0*t*sin\alpha + 1/2*g*t^{2}$

Let´s replace with the given data:

$0 = 22m + 39.3 m/s*t -4.9 m/s^{2}*t^{2}$

Solving the quadratic equation:

t = -0.5 and t = 8.5 s

At 8.5 s after firing, the ball hits the ground.

Now, we can find the module of the velocity vector when the ball hits the ground:

$v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)$

at time t = 8.5s

$v = (81.0m/s * cos\alpha ; 81.0m/s * sin\alpha - 9.8 m/s^{2} *8.5s)$

v = (70.8 m/s; -44.0 m/s)

module of v = $\sqrt{(70.8m/s)^{2} + (-44.0m/s)^{2}}$ = 83.4 m/s